\(D=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
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T
23 tháng 2 2017
\(=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)\)
\(=2002.\frac{5}{24}\)
\(=\frac{5005}{12}\)
PM
2
25 tháng 3 2015
A=202202.1/1212+202202.1/2020+202202.1/3030+202202.1/4242+202202.1/5656
A=202202.(1/1212+1/2020+1/3030+1/4242+1/5656)
A=202202.5/2424
A=417/1/12
25 tháng 3 2015
A=202202.1/1212+202202.1/2020+202202.1/3030+202202.1/4242+202202.1/5656
A=202202.(1/1212+1/2020+1/3030+1/4242+1/5656)
A=202202.5/2424
A=5005/12
HP
8 tháng 8 2016
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)
\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\) \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)
\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)
\(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'
\(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)
\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)
\(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)
\(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)
\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)
T
14 tháng 7 2018
a,3^200 và 2^300
3^200=(3^2)^100=9^100
2^300=(2^3)^100=8^100
Vì 9^100>8^100=>3^200>2^300
Vậy 3^200>2^300
b, 71^50 và 37^75
71^50=(71^2)^25=5041^25
37^75=(37^3)^25=50653^25
Vì 5041^25<50653^25=> 71^50<37^75
Vậy 71^50<37^75
c, 201201/202202 và 201201201/202202202
201201201/202202202=201201/202202
=> 201201/202202=201201201/202202202
Vậy 201201/202202=201201201/202202202
AH
14 tháng 7 2018
a)
Ta có:3200=32.100=(32)100=9100
2300=23.100=(23)100=8100
Vì 9100>8100
Nên 3200>2300
b)
Ta có: 7150=712.25=(712)25=504125
3775=373.25=(373)25=5065325
Vì 504125<5065325
Nên 7150<3775
c)
Ta có:
201201/202202=201.1001/202.1001=201/202
201201201/202202202=201.1001001/202.1001001001= 201/202
Vì 201/202=201/202
Nên 201201/202202=201201201/202202202
DP
10 tháng 7 2017
\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)
\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)
S
1
DS
10 tháng 3 2016
a. 3200 = (32)100 = 9100
2300 = (23)100 = 8100
Vì 9100 > 8100 => 3200 > 2300
NM
5
20 tháng 5 2022
Ta có:\(\dfrac{201201}{202202}\)=\(\dfrac{201}{202}\);\(\dfrac{201201201}{202202202}\)=\(\dfrac{201}{202}\)
=>\(\dfrac{201}{202}\)=\(\dfrac{201}{202}\)
=> \(\dfrac{201201}{202202}\)=\(\dfrac{201201201}{202202202}\)
Vậy: \(\dfrac{201201}{202202}\)=\(\dfrac{201201201}{202202202}\)
(quá dễ)
\(D=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(D=\frac{2002.101}{101.12}+\frac{2002.101}{20.101}+\frac{2002.101}{30.101}+\frac{2002.101}{42.101}+\frac{2002.101}{56.101}\)
\(D=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)
\(D=\frac{1001}{6}+\frac{1001}{10}+\frac{1001}{15}+\frac{143}{3}+\frac{143}{4}\)
\(D=\frac{5005}{12}\)
đáp án là 5005/12 nhé bạn
tích cho mik nha